Question

The mean per capita income is 24,653 dollars per annum with the standard deviation of 778 dollars per annum. What is the probability that the sample mean would be less than $24,745 if a sample of 441 persons is randomly selected? Round your answer to four decimal places

Answer 1

Remember that

`z=(x-μ)/((σ)/(√(n)))`

where

μ=24,653

σ=778

n=441

X=24,745

substitute

`\begin{gathered} z=(24,745-24,653)/((778)/(√(441))) \\ \\ z=2.4833 \end{gathered}`

using the values of the z-score table

we have that

P(x>2.4833) = 0.0065086

therefore

The answer is 0.0065